Question

嗨，我有一个dataframe，其中包含三列：名称列，数字列A和数字列B。我要做的是将行按它们在A列中的值进行装箱，然后找到B列中每个合并组的值的标准偏差。

我目前已经弄清楚了如何将垃圾箱设置成相等的长度间隔

sapply(df, function(x) {

bins <- with(x, cut(x$a, breaks = seq( floor(min(x$a)/10)*10, ceiling(max(x$a)/10)*10, 
        by= 10), include.lowest = T, labels = labels))})

但是我仍然坚持如何将bin应用于数据框并计算B列中每个bin组的一组值的标准偏差。

有人有什么建议吗？谢谢...

Answer 1

jupyter notebook

Answer 2

下面是一个示例，该示例基于合并的b（其中应用了a，即

）来计算ave的{{1}}

df <- within(df, bins <- cut(a, breaks = seq( floor(min(a)/10)*10, ceiling(max(a)/10)*10, by= 10), include.lowest = T))
df <- within(df, b.sd <- ave(b,bins,  FUN = sd))

这样

> df
           a           b     bins      b.sd
1  26.550866 -0.05612874  (20,30] 0.4936856
2  37.212390 -0.15579551  (30,40] 0.5699030
3  57.285336 -1.47075238  (50,60] 1.5255800
4  90.820779 -0.47815006 (90,100] 0.6159115
5  20.168193  0.41794156  (20,30] 0.4936856
6  89.838968  1.35867955  (80,90] 1.0255606
7  94.467527 -0.10278773 (90,100] 0.6159115
8  66.079779  0.38767161  (60,70] 1.3737593
9  62.911404 -0.05380504  (60,70] 1.3737593
10  6.178627 -1.37705956   [0,10] 0.4163140
11 20.597457 -0.41499456  (20,30] 0.4936856
12 17.655675 -0.39428995  (10,20] 0.5249513
13 68.702285 -0.05931340  (60,70] 1.3737593
14 38.410372  1.10002537  (30,40] 0.5699030
15 76.984142  0.76317575  (70,80] 0.7825274
16 49.769924 -0.16452360  (40,50] 1.2127445
17 71.761851 -0.25336168  (70,80] 0.7825274
18 99.190609  0.69696338 (90,100] 0.6159115
19 38.003518  0.55666320  (30,40] 0.5699030
20 77.744522 -0.68875569  (70,80] 0.7825274
21 93.470523 -0.70749516 (90,100] 0.6159115
22 21.214252  0.36458196  (20,30] 0.4936856
23 65.167377  0.76853292  (60,70] 1.3737593
24 12.555510 -0.11234621  (10,20] 0.5249513
25 26.722067  0.88110773  (20,30] 0.4936856
26 38.611409  0.39810588  (30,40] 0.5699030
27  1.339033 -0.61202639   [0,10] 0.4163140
28 38.238796  0.34111969  (30,40] 0.5699030
29 86.969085 -1.12936310  (80,90] 1.0255606
30 34.034900  1.43302370  (30,40] 0.5699030
31 48.208012  1.98039990  (40,50] 1.2127445
32 59.956583 -0.36722148  (50,60] 1.5255800
33 49.354131 -1.04413463  (40,50] 1.2127445
34 18.621760  0.56971963  (10,20] 0.5249513
35 82.737332 -0.13505460  (80,90] 1.0255606
36 66.846674  2.40161776  (60,70] 1.3737593
37 79.423986 -0.03924000  (70,80] 0.7825274
38 10.794363  0.68973936  (10,20] 0.5249513
39 72.371095  0.02800216  (70,80] 0.7825274
40 41.127443 -0.74327321  (40,50] 1.2127445
41 82.094629  0.18879230  (80,90] 1.0255606
42 64.706019 -1.80495863  (60,70] 1.3737593
43 78.293276  1.46555486  (70,80] 0.7825274
44 55.303631  0.15325334  (50,60] 1.5255800
45 52.971958  2.17261167  (50,60] 1.5255800
46 78.935623  0.47550953  (70,80] 0.7825274
47  2.333120 -0.70994643   [0,10] 0.4163140
48 47.723007  0.61072635  (40,50] 1.2127445
49 73.231374 -0.93409763  (70,80] 0.7825274
50 69.273156 -1.25363340  (60,70] 1.3737593

数据

set.seed(1)
df <- data.frame(a = runif(50,0,100),
                 b = rnorm(50))

R根据合并组找到其他色谱柱的标准偏差

2 个答案: